comparisons involving two sample means€¦ · comparison of two sample means (confidence interval)...

COMPARISONS INVOLVING TWO SAMPLE MEANS

Testing Hypotheses

Two types of hypotheses:

1. Ho: Null Hypothesis - hypothesis of no difference.

21 or 021

2. HA: Alternate Hypothesis – hypothesis of difference.

21 or 021

Two-tail vs. One-tail Tests

Two-tail tests have these types of hypotheses:

Ho: 21

HA: 21

Note the presence of the equal signs.

If you reject Ho: 21 , you don’t care which mean is greater.

1 can be either greater than or less than

2 .

One-tail tests my have one of the following types of hypotheses:

Ho: 21 Ho: 21

HA: 21 HA:

21

Note the presence of greater than or less than signs.

If you reject Ho:, you are being specific that 1 can only be on one side

of 2 .

t-Distribution

2/ 2/

Two-tail test One-tail test

Size of the rejection region = 2/ Size of the rejection region =

Types of Errors 1. Type I Error: To reject the null hypothesis when it is actually true.

The probability of committing a Type I Error is .

The probability of committing a Type I Error can be reduced by the investigator

choosing a smaller .

typical sizes of are 0.05 and 0.01.

The probability of committing a Type I Error also can be expressed as a

percentage (i.e. *100 %)

If =0.05, The probability of committing a Type I Error is 5%.

If =0.05, we are testing the hypothesis at the 95% level of confidence.

2. Type II Error: The failure to reject the null hypothesis when it is false.

The probability of committing a Type II Error is .

can be decreased by:

a. Increasing n (i.e. the number of observations per treatment).

b. Decreasing s2.

Increase n

Choose a more appropriate experimental design

Improve experimental technique.

Power of the Test

Power of the test is equal to: 1- .

Defined as the probability of accepting the alternate hypothesis when it is true.

We want the Power of the Test to be as large as possible.

Summary of Type I and Type II Errors

True Situation

Decision Null hypothesis is true Null hypothesis if false

Reject the null hypothesis Type I Error No error

Fail to reject the null hypothesis No error Type II Error

Hypothesis Testing

Summary of Testing a Hypothesis.

1. Formulate a meaningful null and alternate hypothesis.

2. Choose a level of .

3. Compute the value for the test statistic (i.e. t-statistic of F-statistic).

4. Look up the appropriate table value for the test statistic.

5. Formulate conclusions.

a. If the tabular statistic > calculated statistic, then you fail to reject Ho.

b. If the tabular statistic < calculated statistic, then you reject Ho.

Comparison of Two Sample Means (t-test)

Suppose we have two population means of and we want to test the hypothesis:

Ho: 21

HA: 21

This can be done using a t-test where:

21

21

YYs

YYt

Where: 1Y = mean of treatment 1

2Y = mean of treatment 2

21 YY

s

= standard deviation of the difference between two means.

Calculation of 21 YY

s

depends on three things.

1. Do the populations have a common variance (i.e. 2

2

2

1 )?

2. Are the two samples of equal size (i.e. n1 = n2)?

3. Are the observations meaningfully paired?

*Comparisons of Two Sample Means (n1 = n2) and 2

2

2

1

Given the following data:

iY 2

iY

Treatment 1 7 9 10 6 9 7 48 396

Treatment 2 2 5 3 1 5 2 18 68

Determine if the treatments means are significantly different at the 95% level of

confidence.

Step 1. Write the hypothesis to be tested:

Ho: 21

HA: 21

Step 2. Calculate 2

2

2

1 & ss

4.216

6

48396

2

2

1

s

8.216

6

1868

2

2

2

s

Step 2.1 Test to see if the two variances are homogeneous (i.e. Ho: 2

2

2

1 ).

The method of calculating 21 YY

s

will depend on whether you reject or fail to reject

Ho: 2

2

2

1 .

If you fail to reject Ho: 2

2

2

1 , the formula for 21 YY

s

is:

n

ss

p

YY

22

21

where: 2

ps = the pooled variance and

n = the number of observations in a treatment total.

If you reject Ho: 2

2

2

1 , the formula for 21 YY

s

is:

2

2

2

1

2

1

21 n

s

n

ss

YY

where: 2

1s = variance of Treatment 1

2


1n = number of observations for treatment 1


Test Ho: 2

2

2

1 using an F-test where:

2

2arg

Smaller

erLF

If there are large differences between 2

2

2

1 & ss , F will become large and result in

the rejection of Ho: 2

2

2

1 .

We will be testing the hypothesis Ho: 2

2

2

1 at the 95% level of confidence.

This F-test is a two-tail test because we are not specifying which variance is

expected to be larger.

Thus, if you are testing 2/ = 0.05, then you need to use the F-table for α = 0.025

(Appendix Table IV, page 611).

This situation (i.e. testing Ho: 2

2

2

1 ) will be the only one used this semester in

which the F-test is a two-tail test.

When we use an F-test to test Ho: 21 , this will be a one-tail F-test because

the numerator of the test, the variance based on means ( 22

Tr ), is expected to

be larger than the denominator, the variance based on individual observations ( 2

). This one-tail F-test requires use of Appendix Table IV on page 610.

For this problem 167.14.2

8.2F

Step 2.2 Look up table F-value in Appendix Table IV (pages 611).

F

)1(),1(,2/ 21 nn = F alpha value/2; numerator df, denominator df

The table in our text is set up for one-tail tests. Thus, to use the table for a two-

tail test and you want to test at the 0.05 level, you need to look up the value of

0.005 in the table. The area under each of the two-tails is 0.025.

For this problem the table F for F0.05/2;5,5=7.15.

Step 2.3 Make conclusions:

Since the calculated value of F (1.167) is less than the Table-F value (7.15), we fail to

reject Ho: 2

2

2


Therefore, we can calculate 21 YY

s

using the formula:

n

ss

p

YY

22

21

Step 3. Calculate 2

ps

The following formula will work in 21 nn or

21 nn

)1()1(

)1()1(

21

2

22

2

112

nn

snsns p

where: 2


2




If 21 nn , you can calculate 2

ps using the formula

2

2

2

2

12 sss p

For this problem 6.22

)8.24.2(2

ps

Step 4. Calculate 21 YY

s

The following formula will work in 21 nn or

21 nn

)11

(21

2

21 nnss pYY

If 21 nn , you can calculate

21 YYs

using the formula:

n

ss

p

YY

22

21

Where n = the number of observations in a treatment total.


)6.2(221

YY

s

Step 5. Calculate t-statistic

21

21

YYs

YYt

371.5

9309.0/5

9309.0

6

18

6

48

t

Step 6. Look up table t-value.

df= )1()1( 21 nn =(6-1)+(6-1)=10

t.05/2;10df = 2.228

Step 6. Make conclusions.

-2228 2.228 5.371

Since t-calc (5.371) is > t-table (2.228) we

reject Ho: at the 95% level of

confidence.

Thus we can conclude that the mean of

treatment 1 is significantly different than that

of the mean of treatment 2at the 95% level of

confidence.

Comparison of Two Sample Means (Confidence Interval)

The formula for a confidence interval to test the hypothesis: Ho: 21 is:

212

21 )(YY

stYY

Using the data from the previous example:

212

21 )(YY

stYY

07.7

93.2

07.25

)931.0(228.2)38(

2

1

l

l

Comparison of Two Sample Means (F-test)

In conducting tests of significance for: 21

21

:

:

A

o

H

H we actually are conducting an F-

test of variances (i.e. the ratio of two variances.

F= estimate of σ2 based on means

estimate of σ2

based on individuals

=2

22

Tr

When the null hypothesis is rejected, the numerator of the F-test becomes large as

compared to the denominator. This causes the calculated vale for F to become large.

So far in class we have seen two different ways to estimate 2 :

1. nY

/22

We can estimate 2 by n* 2

Y (i.e. variance based on means).

2. We can estimate 2 by calculating 2 directly from individuals (variance based on

individuals).

Since the interval does not include

the value 0, we must reject Ho:

at the 95% level of

confidence.

The estimate of 2

Y approaches 2 only when you fail to reject Ho: because the treatment

effect 2

Tr in the expected mean square for the treatment source of variation ( 22

Tr )

approaches zero.

When Ho: 21 is rejected and the treatment variances are homogenous (i.e. 2

2

2

1 ),

the estimate of 2 based on means will over estimate 2 .

This occurs because the estimate of 2 based on means is affected by differences between

treatment means as well as differences due to random chance.

The estimate of 2 based on individuals is not affected by differences between treatment

means.

Note in the models below that the model for observations based on individuals does not

have a component for treatment.

This can be seen by looking at the linear models.

1. Linear model for observations based on individuals.

iiY

Where: Yi = the ith

observation of variable Y.

population mean.

i = random error

2. Linear model for samples from two or more treatments.

ijiijY

Where: Yij = the jth

observation of the ith

treatment.

population mean.

i = the ith

treatment

ij = random error

We can estimate 2 based on means by calculating a value called the Treatment

Mean Square (TRT MS).

We can estimate 2 based on individuals by calculating a value called the Error

Mean Square.

Given the linear model ijiijY , we can rewrite the components as:

can be rewritten as ..Y

i can be rewritten as ... YY i

ij can be rewritten as .iij YY

SOV Df SS MS F

Among trt (Trt) t-1 2... )( YYr i Trt SS/Trt df Trt MS/Error MS

Within trt (Error) t(r-1) 2.)( iij YY Error SS/Error df

Total tr-1 2..)( YYij

r = number of replicates

t = number of treatments

Example

Using the data from the previous t-test and CI problem

.iY 2

iY

Treatment 1 7 9 10 6 9 7 48 396

Treatment 2 2 5 3 1 5 2 18 68

Y..=66

Step 1. Calculate the Total SS

Total SS=

rt

YYYY

ij

ij ij

2

22..)(

Correction factor

Definition formula Working formula

=(72 + 9

2 + 10

2 + . . . + 2

2) - 66

2/(6*2)

= 464 – 363

= 101

Step 2. Calculate Trt SS

Trt SS=

rt

Y

r

YYYr

ijii

22

.2... )(

Correction factor

Definition formula Working formula

75

363438

12

66)

6

18

6

48(

222

Step 3. Calculate Error SS

Direct Method: Error SS= 26)6

1868()

6

48396()(

222

.2 i

i

j

ijr

YY

Indirect Method: Error SS = Total SS – Trt SS = 101-75 = 26

Step 4. Complete the ANOVA Table

SOV Df SS MS F

Trt (t-1) = 1 75.0 75.0 28.85**

Error t(r-1) = 10 26.0 2.6

Total rt-1 = 11

Step 5. Look up the Table F-value.

Fα; numerator df, denominator df = Fα; treatment df, error df

F0.05; 1,10 = 4.96

F0.01; 1,10 = 10.04

This Error MS is the same value as

sp2 in the t-test.

This is the estimate of (i.e. s2).

*,** Significant at

the 95% and 99%

levels of confidence,

respectively.


0 4.96 10.04 28.85

Relationship Between the t-statistic and the F-statistic

When 2

2

2

1 , t2 = F.

Comparisons of Two Sample Means (21 nn ) and 2

2

2

1


iY 2

iY

Treatment 1 3 9 6 10 28 226

Treatment 2 11 15 9 12 12 13 72 884

Determine if the treatments means are significantly different at the 95% level of confidence.


Ho: 21

HA: 21

Step 2. Calculate 2

2

2

1 & ss

1014

4

28226

2

2

1

s

416

6

72884

2

2

2

s

Since F-calc (28.85) > 4.96 we reject

Ho: at the 95% level of

confidence.

Since F-calc (28.85) > 10.04 we

reject Ho: at the 99% level of

confidence.

Step 2.1 Test to see if the two variances are homogeneous (i.e. Ho:2

2

2

1 ).

2

2arg

Smaller

erLF


10F

Step 2.2 Look up table F-value in Appendix Table IV (page 611).

F

)1(),1(,2/ 21 nn = F alpha value/2; numerator df, denominator df

For this problem the table F for F0.05/2;3,5= 7.76.


Since the calculated value of F (2.5) is less than the Table-F value (7.76), we fail to reject

Ho: 2

2

2


Therefore, we can calculate 21 YY

s

using the formula:

)11

(21

2

21 nnss pYY

Step 3. Calculate 2

ps

)1()1(

)1()1(

21

2

22

2

112

nn

snsns p

where: 2


2




25.6)16()14(

4)16(10)14(2

ps


s

614.16

1

4

125.6)

11(

21

2

21

nnss pYY

Step 5. Calculate t-statistic

21

21

YYs

YYt

098.3

614.1

127

t

Many people don’t line working with negative t-values.

If you are conducting a two-tail test, you can work with the absolute value of t since the

rejection regions are symmetrical about the axis of 0.

Therefore, 098.3098.3

Step 6. Look up table t-value.

Df= )1()1( 21 nn =(4-1)+(6-1)=8

t.05/2;8df = 2.306


-2306 2.306 3.098

Since t-calc (3.098) is > t-table (2.306)

we reject Ho: at the 95% level of

confidence.


treatment 1 is significantly different than

that of the mean of treatment 2 at the

95% level of confidence.

Comparisons of Two Sample Means (n1 = n2 or 21 nn ) and 2

2

2

1


2

iY iY

Treatment 1 15.9 13.8 16.5 28.3 28.2 26.1 17.6 3302.60 146.4

Treatment 2 17.6 16.4 14.1 16.2 16.5 14.1 17.7 1824.32 112.6

Determine if the treatments means are significantly different at the 95% level of confidence.


Ho: 21

HA: 21

Step 2. Calculate 2

2

2

1 & ss

12.4017

7

4.1466.3302

2

2

1

s

18.217

7

6.11232.1824

2

2

2

s

Step 2.1 Test to see if the two variances are homogeneous (i.e. Ho: 2

2

2

1 ).

42.1818.2

12.40F

Step 2.2 Look up table F-value.

F0.05/2;6,6=5.82.


Since the calculated value of F (18.42) is greater than the Table-F value (5.82), we reject

Ho: 2

2

2


Therefore, we have to calculate 21 YY

s

using the formula:

2

2

2

1

2

1

21 n

s

n

ss

YY


s

458.27

18.2

7

12.40

2

2

2

1

2

1

21

n

s

n

ss

YY

Step 4. Calculate t’-statistic

Use t’ because t in this case is not distributed strictly as Student’s t.

21

21'

YYs

YYt

964.1

458.2

7

6.112

7

4.146

'

t

Step 5. Calculate effective degrees of freedom.

Calculation of the effective df allows us to use the t-table to look up values to test t’.

Effective df =

65.6)016.0475.5(

516.36

17

7

18.2

17

7

12.40

7

18.2

7

12.40

11

22

2

2

2

2

2

2

1

2

1

2

1

2

2

2

2

1

2

1

n

n

s

n

n

s

n

s

n

s

We can round the effective degrees of freedom of 6.65 to 7.0.

We then look up the t-value with 7 df.

t.05/2; 7df = 2.365


-2.365 1.964 2.365

Paired Comparisons

Earlier it was stated that calculation of 21 YY

s

depends on three things:

4. Do the populations have a common variance (i.e. 2

2

2

1 )?

5. Are the two samples of equal size (i.e. n1 = n2)?

6. Are the observations meaningfully paired?

We will now discuss the consequences of using paired comparisons.

Since t’-calc (1.964) is < t-table (2.365) we

fail to reject Ho: at the 95% level

of confidence.


treatment 1 is not significantly different

than that of the mean of treatment 2at the

95% level of confidence.

Pairing of observations is done before the experiment is conducted.

Pairing is done to make the tests of significance more powerful.

If members of a pair tend to be large or small together, it may be possible to detect smaller

differences between treatments than would be possible without pairing.

The purpose of pairing is to eliminate an outside source of variation, that existing from pair

to pair.

Calculating the variance of differences rather than the variance of individuals controls the

variation.

Examples where pairing may be useful:

Drug or feed studies on the same animal.

Measurements done on the same individual at different times (before and after type

treatments).

DD

s

D

s

YYt

.2.1

where n

DD

jj YYD 21

n = the number of pairs.

n

s

n

ss DD

D

2

11

2

2

2

212

21

n

n

DD

n

n

YYYY

s

jj

jj

D

Example

Replicates Treatment 1 Treatment 2 Difference (D) Difference2 (D

2)

1 8 12 -4 16

2 12 17 -5 25

3 13 16 -3 9

4 10 16 -6 36

5 5 9 -4 16

48.1 Y 70.2 Y 22D 1022D

Step 1. Calculate Ds

14.115

5

22102

1

22

2

n

n

DD

sD

Step 2. Calculate D

s

510.05

14.1

Ds

Step 3. Calculate t-value

627.8510.0

4.4

510.0

5/22

Ds

Dt

Step 4. Look up Table t-value

df = number of pairs – 1

776.24;2/05.1;2/ tt dfn

Step 5. Make conclusions

-8.627 -2.776 2.776

REVIEW

You might be thinking there are many formulas to remember; however, there are some

simple ways to remember them.

We have talked about four types of variances:

s2 = variance based on individuals

2

Ys = variance based on means

2

21 YYs

= variance of the difference between two means

2

Ds = variance of paired observations

However, there is a relationship between all four variances that can help you in

remembering the formulas:

1/n

s2 2

Ys

2 2/n 2

2

Ds 2

21 YYs

1/n

Since t-calc (-8.627) is < t-table (-2.776)

we reject Ho: at the 95% level of

confidence.


treatment 1 is significantly different than

that of the mean of treatment 2at the 95%

level of confidence.

There are only two formulas for t:

Ys

Yt 0 and

21

21

YYs

YYt

You will need to remember how to calculate Y

s and21 YY

s

.

)1()1(

)1()1(

21

2

22

2

112

nn

snsns p regardless if

21 nn or 21 nn

)11

(21

2

21 nnss pYY

regardless if 21 nn or

21 nn

SAS Commands for the t-Test

t-test with equal variance

options pageno=1;

data ttest;

input trt yield;

datalines;

1 7

1 9

1 10

1 6

1 9

1 7

2 2

2 5

2 3

2 1

2 5

2 2

;

proc sort;

by trt;

run;

proc ttest;

class trt;

title 't-test with equal variance';

run;

t-test with unequal variance

options pageno=1;

data unequ;

input trt yield;

datalines;

1 15.9

1 13.8

1 16.5

1 28.3

1 28.2

1 26.1

1 17.6

2 17.6

2 16.4

2 14.1

2 16.2

2 16.5

2 14.1

2 17.7

;;

proc sort;

by trt;

run;

proc ttest;

class trt;

title 't-test with unequal variance';

run;

Paired t-test

options pageno=1;

data paired;

input a b;

datalines;

8 12

12 17

13 16

10 16

5 9

;;

proc ttest;

paired a*b;

title 'paired t-test';

run;

T-test to compare Treatments 1 and 2

(Equal variance)

The TTEST Procedure

Statistics

Variable Trt N

Lower CL

Mean Mean

Upper CL

Mean

Lower CL

Std Dev Std Dev

Upper CL

Std Dev

Std

Err

Mini-

mum

Maxi-

mum

Yield 1 6 6.37 8 9.62 0.96 1.54 3.79 0.63 6 10

Yield 2 6 1.24 3 4.75 1.04 1.67 4.1 0.68 1 5

Yield Diff (1-2) 2.92 5 7.07 1.12 1.61 2.82 0.93

T-Tests

Variable Method Variances DF t Value Pr > |t|

Yield Pooled Equal 10 5.37 0.0003

Yield Satterthwaite Unequal 9.94 5.37 0.0003

Equality of Variances

Variable Method Num DF Den DF F Value Pr > F

Yield Folded F 5 5 1.17 0.8698

If the variances are homogeneous, then we use this t-

test for comparing the means of the two treatments. If

the Pr>|t| value is 0.05 or less, then we reject the null

hypothesis. Thus, we can conclude the two treatment

means are different.

If the variances are not homogeneous, then we use this

t-test for comparing the means of the two treatments.

e two treatment means are different.

This F-test is testing to see if the variances of the two

treatments are homogeneous. If the Pr>F values is

greater than 0.001, then we fail to reject the

hypothesis that the two error variances are

homogeneous.

T-test to compare Treatments 1 and 2

(Unequal variance)

The TTEST Procedure

trt N Mean Std Dev Std Err Minimum Maximum

1 7 20.9143 6.3344 2.3942 13.8000 28.3000

2 7 16.0857 1.4758 0.5578 14.1000 17.7000

Diff (1-2) 4.8286 4.5991 2.4583

trt Method Mean 95% CL Mean Std Dev 95% CL Std Dev

1 20.9143 15.0559 26.7726 6.3344 4.0819 13.9488

2 16.0857 14.7208 17.4506 1.4758 0.9510 3.2499

Diff (1-2) Pooled 4.8286 -0.5276 10.1848 4.5991 3.2979 7.5918

Diff (1-2) Satterthwaite 4.8286 -1.0471 10.7042

Method Variances DF t Value Pr > |t|

Pooled Equal 12 1.96 0.0731

Satterthwaite Unequal 6.6495 1.96 0.0924

Equality of Variances

Method Num DF Den DF F Value Pr > F

Folded F 6 6 18.42 0.0025

Paired t-test

The TTEST Procedure

N Mean Std Dev Std Err Minimum Maximum

5 -4.4000 1.1402 0.5099 -6.0000 -3.0000

Mean 95% CL Mean Std Dev 95% CL Std Dev

-4.4000 -5.8157 -2.9843 1.1402 0.6831 3.2764

DF t Value Pr > |t|

4 -8.63 0.0010

comparisons involving two sample means€¦ · comparison of two sample means (confidence interval)...

Documents